Carbohydrates And Lipids Semi-final Examination 2019
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Questions and Answers
- 1.
Type the enzyme involved in this reaction: Cholesterol + O2 ----- ? -----> Cholestenone +H2O2
Explanation
Cholesterol oxidase is the enzyme involved in the given reaction. It catalyzes the oxidation of cholesterol in the presence of oxygen, resulting in the formation of cholestenone and hydrogen peroxide. This enzyme plays a crucial role in the metabolism of cholesterol and is commonly found in bacteria and fungi.Rate this question:
- 2.
Select the correct lipoprotein electrophoretic pattern under pH 8.6.
- A.
HDL <-- VLDL <-- LDL <-- Chylomicrons
- B.
HDL <-- LDL <-- VLDL <-- Chylomicrons
Correct Answer
B. HDL <-- LDL <-- VLDL <-- ChylomicronsExplanation
The correct lipoprotein electrophoretic pattern under pH 8.6 is HDLRate this question:
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- 3.
Carbohydrates that yield 2 to 10 monosaccharides units upon hydrolysis.
- A.
Disaccharide
- B.
Oligosaccharide
- C.
Polysaccharide
- D.
Monosaccharide
Correct Answer
B. OligosaccharideExplanation
Oligosaccharides are carbohydrates that yield 2 to 10 monosaccharide units upon hydrolysis. This means that when oligosaccharides are broken down by hydrolysis, they will produce 2 to 10 individual monosaccharide molecules. Disaccharides, on the other hand, only yield 2 monosaccharide units, while polysaccharides yield more than 10 monosaccharide units. Monosaccharides are single sugar units and do not require hydrolysis to break down further. Therefore, the correct answer is oligosaccharide.Rate this question:
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- 4.
What is the reference method for measuring cholesterol? (Do not use comma and please no shortcuts)
Correct Answer
Abell levy and brodie, abell levy & brodie, abell levy and brodie method, abell levy & brodie methodExplanation
The reference method for measuring cholesterol is the Abell levy and Brodie method.Rate this question:
- 5.
The only sugar that does not have an active ketone or aldehyde group, thus can not oxidize or reduce other compounds.
- A.
Cellulose
- B.
Lactose
- C.
Maltose
- D.
Sucrose
Correct Answer
D. SucroseExplanation
Sucrose is the correct answer because it is the only sugar among the options that does not have an active ketone or aldehyde group. Ketone and aldehyde groups are necessary for sugars to undergo oxidation or reduction reactions. Since sucrose lacks these groups, it cannot participate in such reactions and therefore cannot oxidize or reduce other compounds.Rate this question:
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- 6.
Which of the following is incorrect?
- A.
Glucose + Glucose = Maltose
- B.
Glucose + Fructose = Sucrose
- C.
Glucose + Lactose = Galactose
- D.
Glucose + Galactose = Lactose
Correct Answer
C. Glucose + Lactose = GalactoseExplanation
The given answer is incorrect because the reaction between glucose and lactose does not produce galactose. In reality, the reaction between glucose and lactose produces lactose, which is a disaccharide composed of glucose and galactose.Rate this question:
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- 7.
Which one is the divisor in the De Long method (mg/dL)?
- A.
6
- B.
2.2
- C.
5
- D.
2.1
Correct Answer
A. 6Explanation
The correct answer is 6. In the De Long method, the divisor is used to convert the unit of measurement from mg/dL to another unit. Since the question does not provide any specific information about the unit conversion, it is not possible to provide a more detailed explanation. However, based on the options given, 6 is the only number that could potentially be used as a divisor in the De Long method.Rate this question:
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- 8.
Reagent of Folin-Wu method
Correct Answer
phosphomolybdateExplanation
The reagent used in the Folin-Wu method is phosphomolybdate. This reagent is commonly used in the Folin-Wu method to detect and quantify reducing substances, such as sugars and amino acids, in a sample. Phosphomolybdate reacts with the reducing substances in the sample to form a blue color complex, which can be measured spectrophotometrically. Therefore, phosphomolybdate is the correct answer for the reagent used in the Folin-Wu method.Rate this question:
- 9.
Which statement is true concerning how ketone bodies are formed?
- A.
Hydrolysis of fatty acids
- B.
Oxidation of glycerol
- C.
Hydrolysis of glycerol
- D.
Oxidation of fatty acids
Correct Answer
D. Oxidation of fatty acidsExplanation
Ketone bodies are formed through the process of oxidation of fatty acids. During this process, fatty acids are broken down and converted into acetyl-CoA molecules, which can then be used to produce ketone bodies. This occurs when the body's glucose levels are low, and it needs an alternative source of energy. The oxidation of fatty acids allows for the production of ketone bodies, which can be used as fuel by various tissues, including the brain, during times of prolonged fasting or low carbohydrate intake.Rate this question:
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- 10.
What is the name of the colorimetric method for triglyceride measurement? (Do not use shortcuts or symbols)
Correct Answer
van handel and zilversmith, van handel and zilversmith methodExplanation
The colorimetric method for triglyceride measurement is known as the van Handel and Zilversmith method. This method involves the use of a color reagent that reacts with the triglycerides, producing a color change that can be measured spectrophotometrically. By comparing the absorbance of the sample to a standard curve, the concentration of triglycerides in the sample can be determined.Rate this question:
- 11.
The screening & diagnosis of gestational DM is through the performance of what tests?
- A.
FBS & Urinalysis
- B.
FBS & OGTT
- C.
OGTT & Serum Electrophoresis
- D.
AOTA
Correct Answer
B. FBS & OGTTExplanation
The screening and diagnosis of gestational DM (Gestational Diabetes Mellitus) is done through the performance of FBS (Fasting Blood Sugar) and OGTT (Oral Glucose Tolerance Test). FBS measures the blood sugar level after fasting for a certain period of time, while OGTT measures the body's ability to process glucose by monitoring blood sugar levels before and after consuming a glucose drink. These tests help identify if a pregnant woman has gestational DM, which is important for timely management and prevention of complications.Rate this question:
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- 12.
The reducing power of NADpH is required for the protection of the cell from oxidative and free radical damage.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
NADPH is a coenzyme that plays a crucial role in cellular redox reactions. It acts as a reducing agent, donating electrons to other molecules and thereby neutralizing free radicals and reactive oxygen species. This helps protect the cell from oxidative damage, which can lead to various diseases and aging. Therefore, the statement that the reducing power of NADPH is required for cell protection from oxidative and free radical damage is true.Rate this question:
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- 13.
Complete end product of Folin-Wu method
Correct Answer
phosphomolybdenum blueExplanation
The complete end product of the Folin-Wu method is phosphomolybdenum blue. This is a compound that is formed when the Folin-Wu reagent reacts with certain substances, such as reducing agents like glucose or other compounds containing phenolic groups. The phosphomolybdenum blue complex has a characteristic blue color, which can be used to measure the concentration of the substance being tested. This method is commonly used in biochemical assays to determine the presence or concentration of various compounds in a sample.Rate this question:
- 14.
A highly branched structure that permits rapid release of glucose from glycogen stores is composed of:
- A.
Glucose-6-PO4
- B.
Glucose-1-PO4
- C.
UDP-glucose
- D.
Glyceraldehyde-3-PO4
Correct Answer
C. UDP-glucoseExplanation
UDP-glucose is a highly branched structure that permits rapid release of glucose from glycogen stores. UDP-glucose serves as a precursor for glycogen synthesis and is responsible for adding glucose molecules to the growing glycogen chain. When glucose is needed, UDP-glucose can be rapidly converted back to glucose-1-phosphate, which can then be converted to glucose-6-phosphate and released into the bloodstream for energy production. This highly branched structure allows for efficient and quick release of glucose when needed.Rate this question:
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- 15.
How many blood sample/s in total is/are drawn in the 2hr post prandial test procedure?
- A.
1
- B.
2
- C.
3
- D.
4
Correct Answer
B. 2Explanation
In the 2hr post prandial test procedure, a total of 2 blood samples are drawn. This means that the individual will have their blood drawn twice during this test.Rate this question:
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- 16.
The compound that may enter the tricarboxylic acid cycle is Glucose-6-PO4.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
Glucose-6-PO4 cannot directly enter the tricarboxylic acid cycle. Before entering the cycle, glucose-6-PO4 must undergo several enzymatic reactions to be converted into pyruvate, which can then enter the cycle as acetyl-CoA. Therefore, the statement is false.Rate this question:
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- 17.
What special tube should be used for lipid determination?
- A.
Green – Heparin
- B.
Gray – NaF
- C.
Blue – Citrate
- D.
NOTA
Correct Answer
D. NOTA -
- 18.
Polarographic method of glucose assay are based on which principle?
- A.
Nonenzymatic oxidation of glucose
- B.
Rate of oxygen depletion measured
- C.
Chemiluminescence from ATP formation
- D.
Change in electrical potential as glucose is oxidized
Correct Answer
B. Rate of oxygen depletion measuredExplanation
The correct answer is "Rate of oxygen depletion measured." In polarographic method of glucose assay, the rate of oxygen depletion is measured to determine the concentration of glucose. This is based on the principle that glucose can be oxidized in the presence of an enzyme, resulting in the consumption of oxygen. By measuring the rate at which oxygen is depleted, the concentration of glucose can be quantified.Rate this question:
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- 19.
This ketone body has the highest concentration among the other ketone bodies.
- A.
Acetoacetate
- B.
Beta hydroxybutyrate
- C.
Acetone
Correct Answer
B. Beta hydroxybutyrateExplanation
Beta hydroxybutyrate is the correct answer because it is the ketone body that has the highest concentration among the other ketone bodies. Acetoacetate and acetone are also ketone bodies, but beta hydroxybutyrate is the predominant ketone body produced during ketosis and is present in the highest concentration in the blood.Rate this question:
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- 20.
Glucose dehydrogenase method for glucose is specific for β-glucose.
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The glucose dehydrogenase method for glucose is not specific for β-glucose. It is a general method that can be used to measure the concentration of glucose, regardless of its specific form or configuration. Therefore, the statement is false.Rate this question:
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- 21.
Which one does not belong in the laboratory results of a urine that belongs to a person with DM type I?
- A.
High specific gravity
- B.
Increased urine volume
- C.
Increased K+
- D.
Presence of ketone bodies
Correct Answer
C. Increased K+Explanation
In a person with DM type I, the kidneys are unable to properly filter and reabsorb glucose, leading to increased glucose levels in the urine. This causes osmotic diuresis, resulting in increased urine volume. Additionally, the breakdown of fatty acids for energy in the absence of insulin leads to the production of ketone bodies, which can be detected in the urine. Both of these findings are characteristic of DM type I. However, increased levels of potassium (K+) in the urine are not typically seen in DM type I. Therefore, increased K+ does not belong in the laboratory results of a urine that belongs to a person with DM type I.Rate this question:
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- 22.
What is the renal threshold for glucose?
- A.
120 to 160 mg/dL
- B.
160 to 180 mg/dL
- C.
130 to 190 mg/dL
- D.
140 to 160 mg/dL
Correct Answer
B. 160 to 180 mg/dLExplanation
The renal threshold for glucose refers to the blood glucose level at which the kidneys start to excrete glucose into the urine. In a healthy individual, this threshold is typically around 160 to 180 mg/dL. Above this range, glucose spills into the urine as the kidneys are unable to reabsorb all of it. This is an important indicator for diabetes, as elevated levels of glucose in the urine can be a sign of uncontrolled blood sugar levels.Rate this question:
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- 23.
A glucose colorimetric test that has an end product of arsenomolybdenum green
Correct Answer
Nelson-Somogyi, Nelson-Somogyi method, Nelson somogyi, Nelson somogyi methodExplanation
The correct answer is Nelson-Somogyi, Nelson-Somogyi method, Nelson somogyi, Nelson somogyi method. This is because the given information states that the glucose colorimetric test produces an end product of arsenomolybdenum green. The Nelson-Somogyi method is a commonly used colorimetric assay for measuring glucose levels, where the reaction between glucose and arsenomolybdenum reagent produces a green color. Therefore, the Nelson-Somogyi method is the correct answer in this context.Rate this question:
- 24.
Which one is also known as the ketosis prone diabetes mellitus?
- A.
DM type I
- B.
DM type II
Correct Answer
A. DM type IExplanation
DM type I is also known as the ketosis prone diabetes mellitus because it is characterized by a lack of insulin production in the body. This leads to a buildup of ketones, which are acidic byproducts of fat metabolism, in the blood. Ketosis is a state in which the body switches to burning fat for fuel instead of glucose. This can result in the development of diabetic ketoacidosis, a serious condition that can be life-threatening if not treated promptly.Rate this question:
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- 25.
Which of the following does not belong in the lipid profile evaluated in the clinical laboratory?
- A.
Triglyceride
- B.
HDL
- C.
Chylomicrons
- D.
VLDL
Correct Answer
C. ChylomicronsExplanation
Chylomicrons do not belong in the lipid profile evaluated in the clinical laboratory because they are large particles that transport dietary fats from the intestines to the liver. They are not measured directly in a lipid profile as their levels can vary greatly depending on recent dietary intake. The lipid profile typically includes measurements of triglycerides, HDL (high-density lipoprotein), and VLDL (very low-density lipoprotein), which provide information about the levels of different types of lipids in the blood and can help assess the risk of cardiovascular disease.Rate this question:
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- 26.
What is the reason for Kussmaul respiration observed in persons with diabetes?
- A.
Expel hydrogen ions
- B.
Compensation mechanism to balance blood pH
- C.
A and B
- D.
Expel carbon monoxide gas
Correct Answer
C. A and BExplanation
Kussmaul respiration is a compensatory mechanism observed in persons with diabetes to balance blood pH. It involves deep, rapid breathing that helps to expel hydrogen ions, which can lower blood pH. Therefore, the correct answer is A and B.Rate this question:
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- 27.
How is total cholesterol computed in the clinical laboratory?
- A.
TC = TAG + HDL + LDL
- B.
TC = HDL + LDL + VLDL
- C.
TC = VLDL + TAG + HDL
- D.
TC = LDL + VLDL + TAG
Correct Answer
B. TC = HDL + LDL + VLDLExplanation
Total cholesterol (TC) is computed in the clinical laboratory by adding up the values of high-density lipoprotein (HDL), low-density lipoprotein (LDL), and very low-density lipoprotein (VLDL). This is because TC is composed of these three types of lipoproteins, which carry cholesterol in the blood. HDL is considered the "good" cholesterol, as it helps remove excess cholesterol from the blood. LDL is known as the "bad" cholesterol, as it can contribute to the buildup of plaque in the arteries. VLDL carries triglycerides and is a precursor to LDL. Therefore, the correct formula for computing TC is TC = HDL + LDL + VLDL.Rate this question:
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- 28.
The reagent used in the ultracentrifugation technique used for lipoproteins has approximately ____ specific gravity.
- A.
1.063
- B.
1.052
- C.
1.093
- D.
1.035
Correct Answer
A. 1.063Explanation
The reagent used in the ultracentrifugation technique for lipoproteins has a specific gravity of approximately 1.063. This specific gravity is important because it allows for the separation of lipoproteins based on their density. By using a reagent with a specific gravity of 1.063, the lipoproteins can be separated into different fractions based on their density, which is crucial for studying their composition and function.Rate this question:
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- 29.
Which one is not an autoantibody that is related to Diabetes Mellitus type I?
- A.
Insulin autoantibody
- B.
Glutamate dehydrogenase autoantibody
- C.
Islet cells autoantibody
- D.
Tyrosine phosphatase autoantibody
Correct Answer
B. Glutamate dehydrogenase autoantibodyExplanation
Glutamate dehydrogenase autoantibody is not related to Diabetes Mellitus type I. Autoantibodies are antibodies produced by the immune system that mistakenly target and attack the body's own tissues. In type I diabetes, the immune system mistakenly attacks and destroys the insulin-producing cells in the pancreas. Insulin autoantibodies, islet cells autoantibodies, and tyrosine phosphatase autoantibodies are all autoantibodies that have been associated with type I diabetes. However, glutamate dehydrogenase autoantibody is not known to be involved in the development or progression of type I diabetes.Rate this question:
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- 30.
What units are used in the ultracentrifugation technique for lipoproteins?
Correct Answer
Svedverg units, SvedvergExplanation
The units used in the ultracentrifugation technique for lipoproteins are Svedberg units. The Svedberg unit (symbol: S) is a unit of time used in sedimentation analysis. It measures the rate at which particles sediment in a centrifugal field. The ultracentrifugation technique involves subjecting a sample to high centrifugal forces to separate particles based on their size and density. By measuring the sedimentation rate of lipoproteins in Svedberg units, scientists can determine their molecular weight and analyze their composition.Rate this question:
- 31.
Ortho-toluidine method incorporates a reagent to produce schiff’s base:
- A.
Neucoproine
- B.
Dubowski
- C.
Aromatic amines
- D.
Acetic acid
Correct Answer
D. Acetic acidExplanation
The ortho-toluidine method involves the use of a reagent to produce Schiff's base. Among the given options, acetic acid is the most suitable reagent for this purpose. Acetic acid can react with ortho-toluidine to form a Schiff's base, which is a type of compound commonly used in organic synthesis and coordination chemistry. Therefore, acetic acid is the correct answer in this context.Rate this question:
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- 32.
A patient with insulinoma may exhibit dizziness and fainting attributable to hypoglycemia.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Insulinoma is a rare tumor of the pancreas that produces excessive amounts of insulin, leading to low blood sugar levels (hypoglycemia). When blood sugar levels drop too low, it can cause symptoms such as dizziness and fainting. Therefore, it is true that a patient with insulinoma may exhibit these symptoms due to hypoglycemia.Rate this question:
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- 33.
A patient should have an _______ diet of ____ carbohydrates per day for ____ days.
- A.
Unrestricted; 100g; 3
- B.
Restricted; 150g, 5
- C.
Restricted; 100g; 5
- D.
Unrestricted; 150g; 3
Correct Answer
D. Unrestricted; 150g; 3Explanation
The correct answer is "unrestricted; 150g; 3". This means that the patient should have an unrestricted diet, meaning they can eat as much as they want, with a limit of 150g of carbohydrates per day for a period of 3 days. This suggests that the patient does not have any specific dietary restrictions and can consume a normal amount of carbohydrates within the given limit for a short period of time.Rate this question:
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- 34.
The reagent of your answer in the previous question
Correct Answer
arsenomolybdate - 35.
A glucose method that utilizes glacial acetic acid
Correct Answer
O'Toluidine, O-Toluidine, O'Toluidine method, O-Toluidine method, O Toluidine method, O Toluidine, Dubowski, Dubowski methodExplanation
The given answer options are all different ways to refer to the same method of using O'Toluidine or O-Toluidine in glucose testing. This method, also known as the Dubowski method, involves using glacial acetic acid along with O'Toluidine to measure glucose levels.Rate this question:
- 36.
A major lipid that contains an unsaturated steroid alcohol formed from four fused rings and a single side chain tail similar to a fatty acid.
- A.
HDL
- B.
Cholesterol
- C.
TAG
- D.
Phospholipid
Correct Answer
B. CholesterolExplanation
Cholesterol is a major lipid that contains an unsaturated steroid alcohol formed from four fused rings and a single side chain tail similar to a fatty acid. It is an essential component of cell membranes and is involved in various biological processes such as hormone synthesis and digestion. Cholesterol is primarily synthesized in the liver and is also obtained from dietary sources. It is transported in the bloodstream by lipoproteins, with high-density lipoprotein (HDL) being considered "good cholesterol" due to its role in removing excess cholesterol from the arteries.Rate this question:
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- 37.
All but one requires cholesterol as a precursor:
- A.
Cortisol
- B.
Estrogen
- C.
Aldosterone
- D.
Dopamine
Correct Answer
D. DopamineExplanation
Dopamine is the correct answer because it does not require cholesterol as a precursor. Cortisol, estrogen, and aldosterone are all steroid hormones that are synthesized from cholesterol. However, dopamine is a neurotransmitter that is produced from the amino acid tyrosine, not cholesterol.Rate this question:
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- 38.
Which pair is hydrophobic and virtually insoluble to water?
- A.
Phospholipid; cholesterol
- B.
Cholesterol ester; triglyceride
- C.
Triglyceride; phospholipid
Correct Answer
C. Triglyceride; pHospHolipidExplanation
Triglycerides are hydrophobic molecules composed of three fatty acid chains and a glycerol backbone. They are virtually insoluble in water due to their nonpolar nature. Phospholipids, on the other hand, have both hydrophobic and hydrophilic regions. While the hydrophilic head of a phospholipid is soluble in water, the hydrophobic tails are not. Therefore, the pair of triglyceride and phospholipid is hydrophobic and virtually insoluble to water.Rate this question:
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- 39.
Product of the method you answered in the previous question.
Correct Answer
Schiff base, schiff bases - 40.
First enzyme involved in the most commonly used enzymatic method
Correct Answer
Glucose oxidaseExplanation
Glucose oxidase is the first enzyme involved in the most commonly used enzymatic method because it catalyzes the oxidation of glucose to produce gluconic acid and hydrogen peroxide. This enzyme is widely used in various industries, including food and beverage, clinical diagnostics, and biosensor applications. Its ability to specifically act on glucose makes it a valuable tool for measuring glucose levels in blood, detecting glucose in food products, and generating hydrogen peroxide for other enzymatic reactions.Rate this question:
- 41.
What is the C-peptide levels in diabetes mellitus type 2?
- A.
Detectable
- B.
Undetectable
- C.
Absent
- D.
Increased
Correct Answer
A. DetectableExplanation
In diabetes mellitus type 2, C-peptide levels are detectable. C-peptide is a byproduct of insulin production, and in type 2 diabetes, the body still produces insulin, although it may be insufficient or not used effectively. Therefore, C-peptide can still be detected in the blood. This is in contrast to type 1 diabetes, where C-peptide levels are usually undetectable or very low, as there is little to no insulin production.Rate this question:
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- 42.
Which of the following the preferred method for measuring ketone bodies?
- A.
Gerhardt's test
- B.
Sodium nitroprusside test
- C.
Beta-hydroxybutyrate dehydrogenase test
- D.
Acetest
Correct Answer
C. Beta-hydroxybutyrate dehydrogenase testExplanation
The preferred method for measuring ketone bodies is the Beta-hydroxybutyrate dehydrogenase test. This test specifically measures the levels of beta-hydroxybutyrate, which is the most abundant ketone body in the blood. It is considered to be more accurate and reliable compared to other methods such as Gerhardt's test, Sodium nitroprusside test, and Acetest.Rate this question:
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- 43.
What is the end color in the sodium nitroprusside test?
- A.
Brown
- B.
Yellow
- C.
Red
- D.
Purple
Correct Answer
D. PurpleExplanation
The end color in the sodium nitroprusside test is purple. This test is commonly used to detect the presence of sulfite ions in a solution. Sodium nitroprusside reacts with sulfite ions to form a purple-colored complex. Therefore, if the solution turns purple after adding sodium nitroprusside, it indicates the presence of sulfite ions.Rate this question:
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- 44.
At what range of albumin values indicate microalbuminuria?
- A.
< 30 mg/24 hrs
- B.
< 10 to 100 mg/24 hrs
- C.
> 200 mg/24 hrs
- D.
30 to 300 mg/24hrs
Correct Answer
D. 30 to 300 mg/24hrsExplanation
Microalbuminuria is a condition where there is a slightly increased amount of albumin in the urine, indicating early kidney damage. The correct answer of 30 to 300 mg/24hrs indicates the range of albumin values that indicate microalbuminuria. This range suggests that albumin levels as low as 30 mg/24hrs and as high as 300 mg/24hrs can be indicative of microalbuminuria. Values below 30 mg/24hrs or above 300 mg/24hrs would fall outside of this range and may suggest different conditions or normal kidney function.Rate this question:
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- 45.
Coupling enzyme utilized in the most common enzymatic method
Correct Answer
peroxidaseExplanation
Peroxidase is the coupling enzyme utilized in the most common enzymatic method. Coupling enzymes are used in enzymatic reactions to link the reaction of interest with another reaction that produces a detectable signal. In this case, peroxidase is used to couple the reaction of interest with the production of a colored or fluorescent compound. This allows for the measurement or detection of the reaction. Peroxidase is commonly used in various biochemical assays and diagnostic tests due to its high specificity and sensitivity.Rate this question:
- 46.
Which of the following is not a reagent in the precipitation method for HDL?
- A.
Heparin
- B.
Sodium phosphotungstate
- C.
Dextran sulfate
- D.
Citrate manganese
Correct Answer
D. Citrate manganeseExplanation
Citrate manganese is not a reagent in the precipitation method for HDL because this method typically involves the use of heparin, sodium phosphotungstate, and dextran sulfate to precipitate the HDL particles. Citrate manganese is not commonly used in this method and therefore is not a reagent in the precipitation process.Rate this question:
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- 47.
First enzyme used in the reference enzymatic method
Correct Answer
hexokinaseExplanation
Hexokinase is the first enzyme used in the reference enzymatic method. This enzyme is responsible for catalyzing the phosphorylation of glucose to glucose-6-phosphate. It is an essential step in glucose metabolism and is commonly used in various biochemical assays to measure glucose levels. Hexokinase has a high affinity for glucose and is widely distributed in various tissues and organisms. Its role in the reference enzymatic method makes it a crucial enzyme for accurate measurement of glucose concentration.Rate this question:
- 48.
A patient for an oral glucose tolerance test (OGTT) must have a blood extraction for how often after the baseline sample:
- A.
Every hour for 3 hours
- B.
Every 30 minutes for 2 hours
- C.
Every 30 minutes for 2 hours
- D.
Every hour for 2 hours
Correct Answer
D. Every hour for 2 hoursExplanation
For an oral glucose tolerance test (OGTT), the patient needs to have their blood extracted every hour for 2 hours after the baseline sample. This is done to monitor the patient's blood glucose levels over a specific period of time after consuming a glucose solution. The frequency of blood extraction allows healthcare professionals to observe how the body processes and regulates glucose, providing valuable information for diagnosing conditions such as diabetes. Extracting blood every hour for 2 hours strikes a balance between obtaining enough data points to accurately assess glucose levels and minimizing discomfort for the patient.Rate this question:
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- 49.
All of the following are characteristics of severe hyperglycemia, except:
- A.
Polyuria
- B.
Hyperinsulinism
- C.
Ketonuria
- D.
Glycosuria
Correct Answer
B. HyperinsulinismExplanation
Severe hyperglycemia is characterized by high blood sugar levels. Polyuria, ketonuria, and glycosuria are all common symptoms of severe hyperglycemia. Polyuria refers to excessive urination, ketonuria is the presence of ketones in the urine, and glycosuria is the presence of glucose in the urine. However, hyperinsulinism is not a characteristic of severe hyperglycemia. Hyperinsulinism refers to an excessive production or secretion of insulin, which would actually lower blood sugar levels rather than cause hyperglycemia.Rate this question:
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- 50.
Most of this lipid’s subclasses are involved in the structure of the cell membrane
- A.
Phospholipid
- B.
Fatty acid
- C.
Cholesterol
- D.
Triglyceride
Correct Answer
A. pHospHolipidExplanation
Phospholipids are a type of lipid that make up the majority of the cell membrane structure. They have a hydrophilic (water-loving) head and hydrophobic (water-fearing) tails, which allow them to form a bilayer in the cell membrane. This arrangement provides a barrier between the inside and outside of the cell, regulating the passage of molecules and ions. Other lipids like fatty acids, cholesterol, and triglycerides have different functions and are not primarily involved in the structure of the cell membrane.Rate this question:
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